find-0-1-dx-1-x-n-interms-of-digamma-

Question Number 174117 by Mathspace last updated on 25/Jul/22

find ∫_0 ^1 (dx/(1+x^n )) interms of ψ (digamma)

$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}^{{n}} }\:{interms}\:{of} \\ $$$$\psi\:\left({digamma}\right) \\ $$

Answered by Mathspace last updated on 25/Jul/22

∫_0 ^1 (dx/(1+x^n ))=∫_0 ^1 Σ_(p=0) ^∞ (−1)^p x^(np) dx =Σ_(p=0) ^∞ (−1)^p ∫_0 ^1 x^(np) dx =Σ_(p=0) ^∞ (((−1)^p )/(np+1)) =Σ_(k=0) ^∞ (1/(2nk+1))−Σ_(k=0) ^∞ (1/((2k+1)n+1)) =Σ_(k=0) ^∞ ((1/(2nk+1))−(1/(2nk+n+1))) =nΣ_(k=0) ^∞ (1/((2nk+1)(2nk+n+1))) =(n/(4n^2 ))Σ_(k=0) ^∞ (1/((k+(1/(2n)))(k+((n+1)/(2n))))) =(1/(4n))(ψ(((n+1)/(2n)))−ψ((1/(2n))))×(1/(((n+1)/(2n))−(1/(2n)))) =(1/(2n)){ψ((1/(2n))+(1/2))−ψ((1/(2n)))}

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}^{{n}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \sum_{{p}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{p}} {x}^{{np}} {dx} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{p}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{np}} {dx} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{p}} }{{np}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}{nk}+\mathrm{1}}−\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right){n}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{2}{nk}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{nk}+{n}+\mathrm{1}}\right) \\ $$$$={n}\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{nk}+\mathrm{1}\right)\left(\mathrm{2}{nk}+{n}+\mathrm{1}\right)} \\ $$$$=\frac{{n}}{\mathrm{4}{n}^{\mathrm{2}} }\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({k}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\left({k}+\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{n}}\left(\psi\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\right)×\frac{\mathrm{1}}{\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{2}{n}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}}\left\{\psi\left(\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\right\} \\ $$

Commented by aleks041103 last updated on 25/Jul/22

$${Yes},\:{this}\:{is}\:{correct}!\:{I}\:{made}\:{an}\:{extremely}\: \\ $$$${big}\:{mistake}\:{of}\:{doing}\:{whatever}\:{I}\:{want}\:{with} \\ $$$${nonconverging}\:{integrals}… \\ $$

Commented by Mathspace last updated on 25/Jul/22

$${your}\:{method}\:{is}\:{correct}\:{by}\:{you}\:{hsve}\:{commited} \\ $$$${a}\:{error}\:{of}\:{calculus}\:{nevermind}\:{sir} \\ $$

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